EdSim51 - Notes on the 8051

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The 8051

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Peripheral Interfacing

The following topics are covered in this section:

InputsInterfacing Digital Inputs to TTLSwitchesKeypadsADC Interfacing

Outputs

LEDs and 7-segment DisplaysMultiplexing

DC Motors

Inputs

Interfacing Digital Inputs to TTL

Before we can discuss interfacing digital signals to TTL inputs, we must remind ourselves of the voltage and current levels of a TTL input.

The logic levels for TTL inputs are as follows:

HIGH - between 2 V and 5 VLOW - between 0 V and 0.8 V

When a HIGH is applied to a TTL input it draws very little current (about 40 uA). When a LOW is applied to a TTL input it sourcesapproximately 1.6 mA.

Therefore, when interfacing a digital input (which will be either HIGH or LOW) to TTL we must ensure it meets the requirements of a TTLinput.

The diagram below shows how to interface a digital input (ie; either high or low) to TTL.

When the digital input is HIGH the transistor will be turned on. This results in a direct path from the port pin to ground, therefore the pin is

logic 0. When the digital input is LOW the transistor is off which means there is no path for current from the collector to the emitter,therefore the port pin will read 5V.

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This circuit results in logic inversion, but this should not be a problem as inverting the port pin through software is very easy.

When the transistor is on the port pin is connected directly to ground. Therefore there is a path for the 1.6 mA from the port pin (TTLcurrent when input is LOW). The 10 K emitter resistor ensures the current from the supply is kept low, an important consideration inbattery powered devices.

When the transistor is off the port pin is at 5 V (in reality it will not be exactly 5V because a transistor is never fully off and a small currentwill flow through the emitter resistor, resulting in a small voltage drop across the resistor - but the voltage level on the resistor will still beclose to 5 V). Since there is very little current flowing, power consumption is kept low. Also, as mentioned above, a TTL input draws verylittle current when a HIGH is applied to it, so this circuit satisfies the requirements of a TTL input and keeps power consumption at aminimum.

The digital HIGH need not be TTL level inputs.

The HIGH voltage applied to the base of the transistor does not have to be 5V. For example, the input circuit could be from temperaturesensor that produces a HIGH voltage when the temperature exceeds a certain value. The voltage level from this circuit could be any voltagethat is high enough to turn on the transistor (say, 12V for example) and the value of the base resistor can be calculated to suit. Because thevoltage applied to the emitter resistor is 5 V the digital HIGH from the sensor circuit is level-shifted to suit a TTL input.

Switches

The circuit above is actually interfacing an electronic switch, the transistor, to a TTL input. We can interface a physical switch in exactly the

same way, as shown below.

When the switch is open, no current flows through the resistor and therefore the voltage on the microcontroller pin is 5 V. When the switchis closed the pin is connected directly to ground. As before, when the TTL input is HIGH practically no current flows in the circuit and when

the input is LOW there is a direct current for the 1.6 mA that may flow from the pin.

Switch Bounce

When a physical switch is closed the contacts bounce open and closed rapidly for about 20 to 30 ms, as illustrated below.

The opening of a switch is clean and without bounce. However, when a switch is closed the contacts bounce open and closed for about 30ms. While this is a very short time in human terms it is a very long time for a microcontroller (the basic 8051 running on a system clock of 12

MHz executes a 1-byte instruction in 1 us). Without switch debouncing, the microcontroller would 'think' the switch was opened and closedmany times. Imagine if a push-button switch was being used to increment the output from a microcontrolled power supply. If the switch was

connected to the microcontroller without switch bounce then a user pressing the switch once would actually result in the output voltagebeing increased many times because the microcontroller would respond as if the switch had been pressed many times.

Hardware Switch Debounce

One method of hardware debounce is shown below.



As can be seen from the waveforms above, when the switch is closed the capacitor voltage drops to 0 V. When the switch bounces open

the capacitor begins to charge back up to 5 V, but due to the RC constant, it cannot charge beyond the schmitt trigger's HIGH referencevoltage before the switch bounces closed again. Once the switch closes, the capacitor again drops to 0 V. Since the voltage on the

capacitor never goes beyond the trigger's HIGH reference voltage during the switch bounce, the output of the trigger remains LOW and theswitch bounce is not seen on the port pin.

When the switch is physically opened the capacitor has plenty of time to charge beyond the trigger's HIGH reference voltage, at which pointthe trigger output goes HIGH and this is seen on the port pin.

Software Switch Debounce

Hardware debounce requires extra circuitry. With small devices, this extra hardware may not be tolerated. In most cases, software

debounce is more than adequate. Software debounce simply incorporates a delay of about 30 ms while the switch bounces. In other words,when a key press is detected, the system delays for about 30 ms before processing the input. By then the switch will have stopped bouncing

and the microcontroller will only process the initial switch contact.

Delaying for about 30 ms will not be noticed by the user, but the microcontroller could do quite a bit of work in that time. Therefore, in amultifunction system it is more efficient to initialise a timer and start it so that it will overflow and generate an interrupt 30 ms after the switch

was pressed. In that way, the controller can be doing some other work while the switch bounces.



In the above flow chart diagrams, processing the key press is handled in the ISR. This may not be the case in a multifunction system. It ismore likely that the ISR would set a flag to let the main program know that a key has been pressed (or, in an RTOS system, to move aprocess from the waiting state to the ready state).

Keypads

Keypads are assembled in a matrix form, as illustrated below.

The above diagram shows a 4 X 4 keypad - 16 switches configured in 4 columns and 4 rows.

In the default state (all switches open) there is no connection between the rows and columns. When a switch is pressed a connectionbetween the switch's row and the switch's column is made.

Keypad Decoder

Many keypads are built with an onboard decoder that scans the keypad and, if a key is pressed, returns a number identifying the key.Alternatively, a keypad decoder chip can be purchased separately and interfaced with a keypad. The diagram below shows a 4 X 4keypad interfaced with such a decoder.



The decoder is shown connected to the 8051, as a suggestion. Obviously, the outputs from the decoder don't need to go to the LSBs ofP1, nor does the data available line (DA) need to be connected to the external 0 interrupt line. However, the above configuration is oneway of interfacing a 4 X 4 keypad decoder to the 8051.

The decoder has 8 inputs; the 4 X inputs are connected to the 4 keypad columns while the Y inputs are connected to the 4 keypad rows.

Not shown in the diagram are pins for connecting capacitors to the decoder. These capacitors govern the rate at which the keypad isscanned.

When a key is pressed the 4-bit code for the key (there are 16 keys, therefore there are codes 0000 to 1111 in binary) appears on the fouroutput lines (and in this case will appear on the 4 LSBs of P1) and the data available line (DA) goes LOW. If connected to an external

interrupt line (in this case, the INT0-bar line) the microcontroller will be interrupted when a key is pressed. The ISR could then read the 4LSBs of P1 and process the data.

The decoder chip takes care of switch debounce, therefore the programmer is freed from this responsibility, which is an advantage of usinga decoder chip.

Software Decoder

The above decoder function can be implemented in software. The keypad could be interfaced with the 8051 as detailed below.

With the above configuration, an interrupt is generated on the INT0-bar line when a key is pressed. We will deal with how this works in amoment. Firstly, let's see how the keyboard is scanned.

The steps are:

Scan row 1

Scan row 2



Scan row 3Scan row 4

Scanning a row is achieved by applying 0 V to the port pin for that row and 5 V to the other three rows, then scanning each individual

column to see if one of them is LOW. If it is, then the key at the junction between the current row and column being scanned is the pressedkey.

Clear row 1, set other 3Scan column 1

Scan column 2Scan column 3Scan column 4

Clear row 2, set other 3

Scan column 1Scan column 2


Clear row 3, set other 3Scan column 1

Scan column 2


Clear row 4, set other 3


Scan column 3

Scan column 4

For example, let's say the key being pressed is key 6. When scanning the first row, P1.0 will be cleared while the other 3 rows (P1.1, P1.2and P1.3) are set, as detailed in the diagram below.



Since no key in this row is closed there is no path for current through any of the pull-up resistors and all 4 columns (on P1.4 to P1.7) areHIGH. Therefore, the key pressed was not found while scanning row 1.

The diagram below illustrates scanning row 2. (Note that key 6 is still closed.)

In this case, column 3 is connected through the closed switch to row 2. Since row 2 is LOW, column 3 is LOW.



A series of flowcharts for implementing software keypad scan is given below:





The keypad is initialised in the main program; all rows are cleared. Therefore, when a key is pressed, since all rows are LOW, then one of

the columns (the one containing the key that has been pressed) will be connected to 0 V. This logic 0 into the AND gate will result in a logic0 out. Since the output of the AND gate is connected to INT0-bar, a key press will result in an external 0 interrupt.

The keyPressedISR is the ISR for dealing with an external 0 interrupt. The first thing this ISR does is disable the external 0 interrupt and call

a 30 ms (typical) delay subroutine. When the delay subroutine returns the key will have stopped bouncing and the keypad can now be

scanned.

In this example, the value of the key pressed (0 to 15) will be stored in R0. Therefore, at the start of the scanKeypad subroutine R0 is setto 0. Row 1 is cleared while the other three rows are set. The columnScan subroutine is then called. It tests the status of each column. If the

first column is 0 then, since we are currently scanning row 1, key 0 was pressed and its value is in R0. Therefore the keyFound flag is set

and the subroutine returns.

If the first column is not 0, R0 is incremented (it now holds 1) and the next column is tested. If it is 0 then key 1 was pressed and this valueis in R0 - therefore the keyFound flag is set and the subroutine returns.

This is repeated for all columns until the key is found.

If none of the four columns is 0 then the key pressed is not in this row and the subroutine returns without setting the keyFound flag.

When columnScan returns, the keyFound flag is tested. If it is not 1 then the key was not found and the scanKeypad subroutine continueswith the next row. Throughout this process, R0 is incremented so that when the key is finally found R0 will contain the value of that key.

When scanKeypad returns to the ISR the value of the key is in R0. The ISR then calls a subroutine to wait for the key to be released. If wedo not wait for the key to be released and immediately enable the external 0 interrupt, then the system will react as if the key was pressed

again. Even if a user presses and releases the key as quickly as possible, the microcontroller will have executed the ISR and enabled the

INT0 interrupt before the user actually released the key. Therefore we wait for the user to release the key before proceeding.

The waitForKeyRelease subroutine is quite simple. All four rows are cleared so that no matter which key is pressed, one of the columns will

be 0 which will result in a 0 out of the AND gate to the INT0-bar line. Therefore, while the key is held down this line will be 0. Once thekey is released then all four columns will be 1 resulting in a 1 at INT0-bar and the subroutine returns.

The external 0 interrupt flag (IE0) is then cleared because it may have been set as the key bounces and as we scanned the keyboard



(remember, disabling an interrupt does not prevent the interrupt from occurring - a 0 on INT0 will still set the flag, IE0). If we did not clear

this flag, once we again enable the external 0 interrupt the system would again vector to keyPressedISR.

The processKeyFlag is then set to alert the main program to the fact that a key was pressed. The main program can then retrieve the valueof the key from R0 and process it in whatever way the system design requires.

Separate Debounce Interrupt

One of the problems with the above implementation is the fact that the system is held in a 30 ms delay while the key bounces. The

microcontroller could be doing something a lot more useful in that time (remember, a basic 8051 running on a system clock of 12 MHzcould execute 30,000 1-byte instructions in that time). One way would be to set one of the timers to overflow in 30 ms, enable the timer's

interrupt, start the timer and exit the keyPressedISR. The timer's ISR could then execute the rest of the keypad scan process (ie; what's leftin keyPressedISR above goes into the timer's ISR instead).

Doing Nothing While the Key is Held Down

In the above implementation the system is held up doing nothing as it waits for the key to be released. Even in the short space of time of an

ordinary key press, the controller could be doing something useful, but it is even more wasteful if you imagine a user holding the key downfor a long time. One solution with the 8051 would be to initialise the external 0 interrupt as negative-edge activated rather than low-level

activated. In this case, the waitForKeyRelease subroutine would not be needed at all. Once the external 0 interrupt is again enabled in

keyPressedISR another external 0 interrupt could not occur until the key was released and another (or the same one) pressed; only thiswould result in a negative edge on INT0-bar.

Note: there is one other task performed by waitForKeyRelease - in performing its function it resets all four rows. If this subroutine isremoved, as suggested, then the code for clearing the rows (ie; CLR P1.0; CLR P1.1; CLR P1.2; CLR P1.3) must be included in

keyPressedISR so that the system is then ready for the next key press.

ADC Interfacing

The schematic symbol for a typical analogue-to-digital converter is shown below. On the right is an illustration of how the ADC may be

interfaced with the 8051.

The function of the ADC pins are as follows:

CS-bar is the chip selectINTR-bar is the interrupt line - goes low when a conversion is complete.

RD-bar enables the data lines.

WR-bar is cleared and then set to start a conversion.

In the above example of interfacing the ADC to the 8051, the CS-bar line is connected to ground to permanently enable the chip. TheINTR-bar line goes LOW once a conversion is complete, therefore it is connected to one of the external interrupt pins on the 8051. In this

way, the 8051 will be interrupted when a conversion is complete and data is ready for reading.

The data lines are tri-state (hence the inverted triangle symbol) which means this chip can be memory mapped and the data lines can be



connected directly to the data bus. In the above example the data lines are connected to port 1, but since they are tri-state the port can also

be used for something else. Only when the conversion is complete is P2.0 cleared which enables the data lines and the analogue conversion

appears on port 1.

The WR-bar line is used for starting a conversion. Clearing this line resets the internal successive-approximation register and the 8-bit shiftregister. When the line is set conversion begins.

Therefore, taking a reading from the ADC is a two step process:

1. Clear and then set WR-bar to initiate a conversion.

2. Sometime later (typically 100 us), the INTR-bar line will go LOW to indicate the conversion is complete. This will cause an external

0 interrupt and it is up to the external 0 ISR to read the data by clearing P2.0 and reading the data from port 1.

Using an interrupt allows the microcontroller to do some work during the 100 us it takes to convert the analogue input into digital.

Outputs We will now look at interfacing output devices to microcontroller ports, focusing on the following topics:

LEDs

7-segment LED displaysMultiplexing 7-segment LED displays

DC motors

TTL Compatibility

When discussing interfacing to TTL inputs above, we noted the current sinked by a TTL input when the input is HIGH is approximately 40

uA. Since the fan-out of TTL is 10, the maximum current sourced by a TTL output HIGH is 400 uA.

The current sourced by a TTL input when the input is LOW is 1.6 mA, therefore (with a fan-out of 10) the maximum current sinking for a

TTL output LOW is 16 mA.

LEDS

Therefore, when interfacing an LED to a TTL output, the maximum current through the LED is 16 mA. The circuit below shows how to

interface an LED to a microcontroller port pin.

Basic LED interface providing max. LED on current of 16

mA LED interface to provide extra LED on current

The value for the resistor is calculated to result in the desired LEDon current - most LEDs have a forward voltage drop of about 2 V.

The values of both resistors are calculated, using the

transistor Bdc value, to achieve the desired LED on

current.

As explained, this limits the LED on current to 16 mA, which in most cases is adequate. However, if more current is required, the second

circuit shown above may be used. Note that both circuits result in the microcontroller port pin sinking current when the LED is on, which is

desirable as port pins will sink a lot more current than they will source.

7-segment Displays



Interfacing a single 7-segment display to a microcontroller port is done in the same manner as interfacing a single LED. Again, the LEDs canbe connected directly to the port pins or, if high current LEDs are being used, they can be connected through p-n-p transistors. Both

methods are detailed below.

Basic 7-segment display interface providingmax. LED-on current of 16 mA

7-segment interface to provide extra LED-on current

The complete configuration for only one segment and the decimal point

is shown. However, the configuration for the other 6 segments is exactly

the same.

Multiplexing

It is often necessary to interface a number of 7-segment displays to a microcontroller. Rather than use a separate port for each display, allof the displays are connected to the same port and other port pins are used for switching on one display at a time.

As long as the displays are turned on and off fast enough, to the eye it will seem as if they are all on together. A timer interrupt could be

used to switch from one display to the next. For example, timer zero could be set to interrupt once every five milliseconds. The operation of

the timer zero ISR would be:

Turn off all displays by setting the display enable port pins.

Send data for next display to the data port.

Turn on next display (if current display is 4, next display is 1) by clearing the appropriate enable port pin.



DC Motors

An interfacing method for turning on and off a DC motor via a microcontroller is shown below.

However, the above circuit will only work for a 5 V motor. If the supply voltage is changed (for example, if the supply is changed to 12 V

to run a 12 V motor) then the motor will be on all the time because 5 V applied to the base of the p-n-p transistor is not enough to turn it

off.

To interface to larger motors the following circuit may be used.

In the above example a 12 V DC motor is interfaced to a microcontroller. When the port pin is set (ie; is equal to 5 V) the p-n-p transistor

is off which means the n-p-n transistor is also off. Therefore there is no path for current through the motor and the motor is off.

When the port pin is cleared the p-n-p transistor is on. This turns on the n-p-n transistor which allows current to flow through the motor to

ground; the motor is on.

The value of R2 needs to be carefully chosen; too high and the current into the base of the n-p-n transistor will not be enough to turn on the

transistor, too low and the circuit draws too much current.

Bi-directional DC Motor

A circuit diagram for interfacing a 12V DC motor to a microcontroller in a way that allows the controller to not only turn on and off the



motor but also to set the direction in which the motor runs when it is on, is given below.

The circuit is made up of a bridge. If both sides of the motor are at the same voltage the motor is off. So, if T1 and T3 are on, both sides of

the motor are connected to 12 V and the motor is off. If T2 and T4 are on both sides of the motor are connected to ground and, again, the

motor is off.

If T1 and T4 are on then the left side of the motor is at 12 V and the right side is at ground, therefore the motor runs in one direction. Wewill call this forward.

If T3 and T2 are on then the left side of the motor is at ground and the right side is at 12 V, therefore the motor runs in the opposite

direction; ie, reverse.

The circuit is designed so that T1 and T2 cannot be on at the same time and T3 and T4 cannot be on at the same time. This is veryimportant; if T1 and T2 were on at the same time there would be a short circuit between 12 V and ground and the transistors would burn

out. The same is true for T3 and T4.

The truth table for the circuit with its two inputs, A and B, is given below.

A B Motor

0 0 off

0 1 (5 V) reverse

1 (5 V) 0 forward

1 (5 V) 1 (5 V) off

An explanation of the four entries in the table is given below:

Entry 1:

With A at 0 (ie; ground) T5 is on which turns on T2; left side of motor is at ground. A at 0 also means T6 is off. Since there is

no path for current through R3 and R4 there is no voltage drop across them which in turn means the base of T1 is at 12 V,

hence it is off.

The right-hand side of the circuit is a mirror image of the left, therefore with B at 0, T4 is on and T3 is off - hence the right sideof the motor is also at ground; the motor is off.

Entry 2:

A is still at 0 which means T1 is still off and T2 is still on; the left side of the motor is at ground.

With B at 5 V (ie; logic 1 on the port pin which is being used for B) T7 is off which means T4 is off. But T8 is on which

generates a path for current through T8 to ground and also through R9 to the base of T3. There is a certain amount of voltage



dropped across R9, but the base of T3 is close enough to ground for T3 to turn on; the right side of the motor is at 12 V.

The motor is therefore on and we stated above that ground on the left of the motor and 12 V on the right would be called

reverse.Entry 3:

This is the mirror image of entry 2, resulting in T1 on, T2 off, T3 off and T4 on; hence the left side of the motor is at 12 V and

the right side is at ground - the motor runs forward.

Entry 4:

As in entry 3, with A at 5 V the left side of the motor is at 12 V.

As in entry 2, with B at 5 V the right side of the motor is at 12 V.Therefore the motor is off.

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Copyright (c) 2005-2006 James Rogers

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EdSim51 - Notes on the 8051